Conditional Probabilities at a Fixed Value

Why Conditioning on a Fixed Value Changes the Calculation

When two measurements are jointly normal and associated, observing one of them changes the distribution we should use for the other. For instance, a high value of $Y$ may make a high value of $X$ more plausible when their correlation is positive, and less plausible when their correlation is negative. To calculate a probability such as $P(X>c\mid Y=y)$, first find the distribution of $X$ conditional on $Y=y$, then standardize $c$ using that distribution.

The conditioning value is fixed, not an interval. In particular, if $Y$ is continuous then $P(Y=y)=0$, so the elementary ratio $P(X>c,Y=y)/P(Y=y)$ is not a usable calculation. Conditional distributions for continuous variables are defined through conditional densities (or, equivalently here, through the bivariate normal model). This gives a well-defined conditional normal distribution at each fixed $y$.

Definition: Suppose $(X,Y)$ has a nondegenerate bivariate normal distribution with $E[X]=\mu_X$, $E[Y]=\mu_Y$, standard deviations $\sigma_X>0$, $\sigma_Y>0$, and correlation $-1<\rho<1$. For a fixed value $Y=y$, the conditional distribution is $$ X\mid Y=y\sim N\left(\mu_X+\rho\frac{\sigma_X}{\sigma_Y}(y-\mu_Y),\ \sigma_X^2(1-\rho^2)\right). $$ The second parameter in this notation is a variance. Thus the conditional mean is $m(y)=\mu_X+\rho\frac{\sigma_X}{\sigma_Y}(y-\mu_Y)$, and the conditional standard deviation is $s=\sigma_X\sqrt{1-\rho^2}$.

Standardizing a Conditional Exceedance

Let $\Phi$ denote the standard normal cumulative distribution function. Once the conditional mean $m(y)$ and standard deviation $s$ are known, standardize the threshold $c$ as $z=(c-m(y))/s$. The probability of exceeding $c$, conditional on $Y=y$, is the upper tail beyond this standardized value.

P(X>c\mid Y=y) =1-\Phi\left( \frac{c-\mu_X-\rho\frac{\sigma_X}{\sigma_Y}(y-\mu_Y)} {\sigma_X\sqrt{1-\rho^2}} \right).

The same standardization also gives lower-tail probabilities: $P(X\le c\mid Y=y)=\Phi((c-m(y))/s)$. Since a nondegenerate conditional normal distribution is continuous, $P(X=c\mid Y=y)=0$; consequently, replacing $>$ with $\ge$, or $\le$ with $<$, does not change the probability.

Find the conditional mean.
Use the observed value $y$ in $m(y)=\mu_X+\rho(\sigma_X/\sigma_Y)(y-\mu_Y)$. Keep the sign of $\rho$ and of $y-\mu_Y$.

Find the conditional standard deviation.
Calculate $s=\sigma_X\sqrt{1-\rho^2}$. The conditional variance does not depend on the particular value $y$.

Standardize the event boundary.
For $X>c$, calculate $z=(c-m(y))/s$; retain the upper tail $1-\Phi(z)$.

Check direction and scale.
If the threshold is above the conditional mean, the exceedance probability must be below $0.5$; if it is below the mean, it must be above $0.5$.

Why the Conditional Normal Formula Holds

A short density argument explains both the conditional mean and the variance adjustment. Define standardized variables $Z_X=(X-\mu_X)/\sigma_X$ and $Z_Y=(Y-\mu_Y)/\sigma_Y$. For a bivariate normal pair with $-1<\rho<1$, its joint density can be factored into the marginal density of $Z_Y$ and a normal density for $Z_X$ given $Z_Y=z_y$:

f_{Z_X,Z_Y}(z_x,z_y) =\phi(z_y)\frac{1}{\sqrt{1-\rho^2}} \phi\left(\frac{z_x-\rho z_y}{\sqrt{1-\rho^2}}\right),

where $\phi$ is the standard normal density. Dividing this joint density by the marginal density $f_{Z_Y}(z_y)=\phi(z_y)$ leaves the conditional density. Therefore, given $Z_Y=z_y$, $Z_X$ is normal with mean $\rho z_y$ and variance $1-\rho^2$. Substituting $z_y=(y-\mu_Y)/\sigma_Y$ and converting back to the units of $X$ gives

E[X\mid Y=y]=\mu_X+\rho\frac{\sigma_X}{\sigma_Y}(y-\mu_Y), \qquad \operatorname{Var}(X\mid Y=y)=\sigma_X^2(1-\rho^2).

The density factorization also shows why the observed value $y$ shifts the conditional mean but does not change the conditional variance in this model. These formulas depend on the bivariate normal assumption; knowing only the means, variances, and correlation of an arbitrary pair is not enough to determine its conditional distribution.

Worked Example: Positive Correlation and an Upper-Tail Probability

A hypothetical monitoring system records two jointly normal sensor readings $X$ and $Y$. Assume $X\sim N(100,15^2)$, $Y\sim N(50,10^2)$, and $\operatorname{Corr}(X,Y)=0.6$. Calculate $P(X>121\mid Y=60)$. There is no interest-rate or mortality assumption in this probability model.

Method. Find the conditional distribution of $X$ at $Y=60$, then use its upper tail. The standardized observed value of $Y$ is $(60-50)/10=1$. Because the correlation is positive, this observation should raise the conditional mean of $X$.

The conditional mean is

m(60)=100+0.6\frac{15}{10}(60-50) =100+0.9(10)=109.

The conditional standard deviation is

s=15\sqrt{1-0.6^2} =15\sqrt{0.64} =15(0.8)=12.

Thus $X\mid Y=60\sim N(109,12^2)$. Standardize the threshold:

z=\frac{121-109}{12}=1.

Therefore,

P(X>121\mid Y=60)=1-\Phi(1)=1-0.8413=0.1587.

Check. The threshold $121$ is one conditional standard deviation above the conditional mean $109$, so a probability near $0.16$ is sensible. The answer, to four decimal places, is $0.1587$. The common mistake of using the marginal distribution would give $1-\Phi((121-100)/15)=1-\Phi(1.4)=0.0808$, which ignores the information in $Y=60$ and is not the requested conditional probability.

Worked Example: Negative Correlation and a High Observation

For a hypothetical pair of jointly normal measurements, let $X\sim N(40,8^2)$, $Y\sim N(20,5^2)$, and $\operatorname{Corr}(X,Y)=-0.5$. Find $P(X>45\mid Y=24)$. No financial interest or mortality assumption is involved.

Method. Apply the conditional normal formula, taking care to preserve the negative sign of the correlation. Here $Y=24$ is $4$ units above its mean. With negative correlation, that raises $Y$ while shifting the conditional mean of $X$ downward.

Calculate the conditional mean:

m(24)=40+(-0.5)\frac{8}{5}(24-20) =40-0.8(4)=36.8.

Calculate the conditional standard deviation:

s=8\sqrt{1-(-0.5)^2} =8\sqrt{0.75} \approx 6.9282.

The threshold $45$ is standardized relative to this conditional distribution:

z=\frac{45-36.8}{6.9282} \approx 1.1830.

Hence

P(X>45\mid Y=24)=1-\Phi(1.1830)\approx 1-0.8816=0.1184.

Check. The threshold is above the conditional mean, so the answer should be below $0.5$; $0.1184$ has the correct direction and scale. Reported to four decimal places, the probability is $0.1184$. Replacing $\rho=-0.5$ by $+0.5$ would incorrectly move the conditional mean upward to $43.2$, reversing the effect of the observed $Y$.

Worked Example: A More Distant Threshold

Suppose hypothetical jointly normal variables $X$ and $Y$ satisfy $X\sim N(200,20^2)$, $Y\sim N(30,6^2)$, and $\operatorname{Corr}(X,Y)=0.75$. Find $P(X>230\mid Y=34)$. The model involves no interest-rate or mortality assumption.

Method. First condition on $Y=34$, which is $4$ units above its mean, and then express the threshold $230$ in conditional standard deviation units.

The conditional mean is

m(34)=200+0.75\frac{20}{6}(34-30) =200+0.75\left(\frac{10}{3}\right)(4) =200+10=210.

The conditional standard deviation is

s=20\sqrt{1-0.75^2} =20\sqrt{0.4375} \approx 13.2288.

Then

z=\frac{230-210}{13.2288} \approx 1.5119, \qquad P(X>230\mid Y=34)=1-\Phi(1.5119)\approx 0.0653.

Check. The threshold is about $1.51$ conditional standard deviations above the conditional mean, so an upper-tail probability near $0.065$ is reasonable. To four decimal places, the answer is $0.0653$. The calculation avoids using the marginal standard deviation $20$, which would produce a different standardized threshold and answer.

Conditions, Interpretation, and Exam Traps

The conditional formula is for a nondegenerate bivariate normal pair: both marginal standard deviations are positive and $-1<\rho<1$. If $\rho=1$ or $\rho=-1$, the pair lies exactly on a line and the conditional variance $\sigma_X^2(1-\rho^2)$ is zero. In those boundary cases, the conditional value is determined by the linear relationship; do not use a positive-standard-deviation normal tail calculation.

At a fixed $y$, the conditional mean is the value predicted by the bivariate normal regression line, while the conditional standard deviation describes the remaining spread around that value. The probability is computed from this conditional distribution, not from the marginal distribution of $X$, and not from the joint probability of an event involving a zero-probability point.

Using the wrong standard deviation. Use $\sigma_X\sqrt{1-\rho^2}$, not $\sigma_X$, for the conditional standard deviation.
Dropping a sign. The shift in the mean is $\rho(\sigma_X/\sigma_Y)(y-\mu_Y)$. A negative correlation or an observation below $\mu_Y$ changes the direction of the shift.
Taking the wrong tail. For $X>c$, calculate $1-\Phi(z)$. For $X\le c$, calculate $\Phi(z)$.
Applying the formula without a normal model. The correlation alone does not determine a conditional distribution for a general pair of random variables.

A useful final check is to compare the threshold with the conditional mean before reading a table or using a calculator. If $c=m(y)$, the exceedance probability is $0.5$. If $c>m(y)$, it is less than $0.5$; if $c<m(y)$, it is greater than $0.5$. This check often catches a sign error or a mistaken tail.

The method is aligned with the SOA Exam P (Probability) syllabus: identify the correct conditional distribution, preserve the model parameters and signs, and standardize relative to the distribution specified by the conditioning information.

Check Your Understanding

For each nondegenerate bivariate normal model below, calculate the requested conditional probability or threshold. Give probabilities to four decimal places unless the question asks for a threshold.

Let $X\sim N(12,3^2)$, $Y\sim N(8,4^2)$, and $\operatorname{Corr}(X,Y)=0.5$. Find $P(X>15\mid Y=12)$. (Answer: $0.2819$)
Let $X\sim N(70,10^2)$, $Y\sim N(100,20^2)$, and $\operatorname{Corr}(X,Y)=-0.4$. Find $P(X>75\mid Y=80)$. (Answer: $0.4566$)
Let $X$ and $Y$ each have standard normal marginal distributions, with correlation $0.8$. Find $P(X>1.5\mid Y=0.5)$. (Answer: $0.0334$)
Let $X\sim N(50,8^2)$, $Y\sim N(30,5^2)$, and $\operatorname{Corr}(X,Y)=0.6$. Find $t$ such that $P(X>t\mid Y=35)=0.10$. Use the standard normal 90th percentile $1.2816$; report $t$ to the nearest cent. (Answer: $63.00$)
Let $X\sim N(120,15^2)$, $Y\sim N(40,8^2)$, and $\operatorname{Corr}(X,Y)=-0.6$. Find $P(X>135\mid Y=44)$. (Answer: $0.0521$)

What You'll Learn